Question

Find dy/dx for y= x^3 ln (cot x)

Answer 1

`(dy)/(dx) = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)`

Step-by-step explanation

By the product rule
`(d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)`, we have

`\begin{aligned}(dy)/(dx) &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}`

By the chain rule:

`\begin{aligned}\big(\ln (\cot x)\big)' &= (1)/(\cot x) \cdot (\cot x)' \\ &= (1)/(\cot x) \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - (\sin x)/(\cos x) \cdot (1)/(\sin^2 x) = - (1)/(\cos x) \cdot (1)/(\sin x) \\&= -\csc(x)\sec(x)\end{aligned}`

By the power rule:

`(x^3)' = 3x^2`

thus

`\begin{aligned}(dy)/(dx) &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}`

Nothing to do to simplify any further, other than factoring out
`x^2`.